Ordinarily, given enough time and trial-and-error(threads), I can get through the very hard puzzles, but the March 23rd is giving me a headache.
So far: R5C4(6), R5C5(9), R5C6(1), and R9C5(3), and THAT's IT!. I need one little break-through. Not the solution, but some logic that'll get me one more number. Any help greatly appreciated.

If you look closely at row 9 and at column 2 you'll see the pair of values {1, 9} lying outside of the bottom left 3x3 box. The way these intersect you can then notice that the pair {1, 9} must occupy the two cells r7c1 & r8c3 -- there's no other way to fit those values in that 3x3 box.

Now there's only one cell left (r7c2) that can hold a "7". And since the only possibilities for r7c9 are {1, 9} you can make more progress in row 7. dcb